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If someone have $4$ red bricks and $8$ blue bricks and arranges them randomly in a circle, what is the probability that two red bricks are not side by side?

Is the sample space $\frac{12!}{4!\,8!} \,=\, 495$ ?

and the answer $P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$ ?

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2 Answers 2

up vote 3 down vote accepted

Suppose there $r$ red bricks, and $b$ blue bricks. Counting clock-wise, let $$ (k_1, k_2, \ldots, k_r) $$ be the tuples, representing the number of blue bricks between two red bricks. The number of favorable configurations is: $$ N_F = \sum_{k_1=1}^{b-r+1} \sum_{k_2=1}^{b-r+1} \cdots \sum_{k_r=1}^{b-r+1} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{t-t^{b-r+2}}{1-t} \right)^r = [t^{b-r}] \left( \frac{1-t^{b-r+1}}{1-t} \right)^r = [t^{b-r}] \left(1-t\right)^{-r} = \binom{b-1}{r-1} $$ and the number of total configurations: $$ N_T = \sum_{k_1=0}^{b} \sum_{k_2=0}^{b} \cdots \sum_{k_r=0}^{b} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{1-t^{b+1}}{1-t} \right)^r= [t^b] \left(1-t\right)^{-r} = \binom{b+r-1}{b} $$ Hence the probability: $$ p =\frac{N_F}{N_T} = \frac{\binom{b-1}{r-1}}{\binom{b+r-1}{b}} = \frac{b! (b-1)!}{(b-r)! (b+r-1)!} $$ With $b=8$ and $r=4$, we have $N_F = \binom{7}{3} = 35$ and $N_T = \binom{11}{8} = 165$, thus $$ p = \frac{N_F}{N_T} = \frac{35}{165} = \frac{7}{33} $$

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I like this solution very much! –  Byron Schmuland Aug 17 '12 at 12:52
    
Presumably every $[t]^a$ should read $[t^a]$, and $[t^{b-1}]$ in the computation of $N_F$ should read $[t^{b-r}]$. But I do not understand why the $t^{b-r-1}$ term in $(1-t^{b-r-1})^r$ should not contribute to $N_F$ as well. (And +1.) –  Did Aug 18 '12 at 21:12
    
@did Thanks for the upvote and comments, and for pointing out the typos. The upper bound in the sums for $N_F$ were incorrect, it should be $b-r+1$ instead of $b-r-1$ (one can keep them to be $+\infty$ to simplify the argument). –  Sasha Aug 19 '12 at 2:00

There are ${12\choose 4}=495$ ways to choose the 4 places to put the red bricks. From my answer here, the number of ways to choose 4 non-consecutive places is ${12\over 8}{8\choose4}=105.$ (Put $d=4$ and $n=12$). The desired probability is then $105/495=7/33=.2121$.

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+1. Please also see my derivation. –  Sasha Aug 17 '12 at 5:07

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